Geometry: Concurrent lines and concyclic points.

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Geometry: Concurrent lines and concyclic points.

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Greece team selection test problem 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

402 posts

Borbas

#1 h May 3, 2017, 11:52 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle inscribed in circle $c(O,R)$ with $AB<AC<BC$ ,
and $c_1$ be the inscribed circle of $ABC$ which intersects $AB, AC, BC$ at
$F, E, D$ respectivelly. Let $A', B', C'$ be points which lie on $c$ such that the quadrilaterals
$AEFA', BDFB', CDEC'$ are inscribable.
(1) Prove that $DEA'B'$ is inscribable.
(2) Prove that $DA', EB', FC'$ are concurrent.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

158 posts

#3 h May 3, 2017, 12:58 PM • 2 Y

Y by Adventure10, Mango247

Easy problem. My solution : $(1)$ $\angle A'ED+\angle A'B'D=\angle A'EF+\angle FED+\angle A'B'F+\angle FB'D=\angle A'AF+\angle A'B'F+\angle B+90-\angle \tfrac{B}{2}=360-((\angle AA'B'=180-\angle B'BF)+(\angle AFB'=180-\angle B'FB)) +90 +\angle \tfrac{B}{2}=90-\angle \tfrac{B}{2}+90+\angle \tfrac{B}{2}=180$ so done!
$(2)$ by radical center of circles $DEA'B' , FEB'C' , FDA'C'$ done!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

8857 posts

#4 h Sep 4, 2022, 8:35 PM

Y by

Write $\measuredangle A'B'D = \measuredangle A'B'B+\measuredangle BB'D = \measuredangle A'AB + \measuredangle BFD = \measuredangle A'EF + \measuredangle FED = \measuredangle AED,$ so we have the first concyclity. The second part simply follows from radical axis.

Edit: Wait, this is just Canada 2007/5. Wonder why it felt familiar.

This post has been edited 1 time. Last edited by HamstPan38825, Sep 4, 2022, 8:39 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

13 posts

#5 h Mar 4, 2025, 5:33 AM

Y by

I think its something wrong with the problem
Its obvious that DEA'B' is cyclic since all four points lie on incircle but in problem it said that it must be inscribable or DE+A'B'=DB'+EA' so i can not understand

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a